Question

Consider the circuit shown where $C_1=6\mu F,C_2=3\mu F$ and $V=20V$. Capacitor $C_1$ is first charged by closing the switch $S_1$. Switch $S_1$ is then opened, and the charged capacitor is connected to the uncharged capacitor $C_2$ by closing $S_2$.

• A. Total heat produced after closing switch $S_2$ is $1.8mJ$.
• B. Total charge that has flown through the battery is $120\mu C$.
• C. Final charge on $C_1$ after opening switch $S_1$ and closing switch $S_2$ is $80\mu C$
• D. Final charge on $C_2$ after opening switch $S_1$ and closing switch $S_2$ is $40\mu C$

Answer 1

The correct option is D Final charge on $C_2$ after opening switch $S_1$ and closing switch $S_2$ is $40\mu C$

When $S_1$ is closed and $S_2$ is open,
charge on $C_1,q_1$ $=C_{1}V=\left(6\mu F\right)\left(20V\right)$
$=120\mu C$
$C_2$ remains uncharged as it is not connected to cell initially

When $S_1$ is open and $S_2$ is closed, after connecting of switch $S_2$ , let charge $q$ flows in circuit as shown.

After connecting of $S_2$, potential difference across $C_1$ and $C_2$ will be the same.

$\frac{120\mu C-q}{C_1}=\frac{q}{C_2}$
($v=$ charge / voltage )

$\frac{120\mu C-q}{6\mu F}=\frac{q}{3\mu F}$

$q=40\mu C$

Thus , charge on $C_1$ and $C_2$ will be $80\mu C$ and $40\mu C$ respectively.
Initial energy stored $\left(E_i\right)=\frac{1}{2}\frac{Q^2}{C_1}$

$=\frac{1}{2}\times \frac{(120\times {10}^{-6}C{)}^2}{6\times {10}^{-6}F}$

Final energy stored $=\frac{1}{2}\frac{Q_1^2}{C_1}+\frac{1}{2}\frac{Q_2^2}{C_2}$

$=\frac{1}{2}\times \frac{(80\times {10}^{-6}C{)}^2}{6\times {10}^{-6}F}+\frac{1}{2}\times \frac{(40\times {10}^{-6}C{)}^2}{3\times {10}^{-6}F}$

∴ Heat produced $=E_i-E_f$ $=\frac{1}{2}\times \frac{(120\times {10}^{-6}C{)}^2}{6\times {10}^{-6}F}-(\frac{1}{2}\times \frac{(80\times {10}^{-6}C{)}^2}{6\times {10}^{-6}F}+\frac{1}{2}\times \frac{(40\times {10}^{-6}C{)}^2}{3\times {10}^{-6}F})$

$=\frac{{10}^{-6}}{2}[2400-\frac{6400}{6}-\frac{1600}{3}]$

$=400\times {10}^{-6}J$
Hence, option (C) is correct.